[编程题] 扫描透镜(本题还涉及如何从字符串中提取数字)

在N*M的草地上,提莫种了K个蘑菇,蘑菇爆炸的威力极大,兰博不想贸然去闯,而且蘑菇是隐形的.只 有一种叫做扫描透镜的物品可以扫描出隐形的蘑菇,于是他回了一趟战争学院,买了2个扫描透镜,一个 扫描透镜可以扫描出(3*3)方格中所有的蘑菇,然后兰博就可以清理掉一些隐形的蘑菇. 问:兰博最多可以清理多少个蘑菇?

输入描述:

第一行三个整数:N,M,K,(1≤N,M≤20,K≤100),N,M代表了草地的大小;
接下来K行,每行两个整数x,y(1≤x≤N,1≤y≤M).代表(x,y)处提莫种了一个蘑菇.
一个方格可以种无穷个蘑菇.

输出描述:

输出一行,在这一行输出一个整数,代表兰博最多可以清理多少个蘑菇.

java

import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class ex3{
    ex3(){}
    public static void main(String[] argvs){
        Scanner scanner = new Scanner(System.in);
        String str = scanner.nextLine();
    
        //从字符串中提取数字
        String regEx = "\\d{1,}";
        Pattern p = Pattern.compile(regEx);
        Matcher m = p.matcher(str);
        int iii = -1;
        int[] t = new int[3];
        while(m.find()){
            iii++;
            if(iii>2)
                break;
            t[iii] =new Integer(m.group()).intValue();
        }
        int N =t[0]; 
        int M =t[1]; 
        int K =t[2]; 
        int x,y;
        int[][] map = new int[N+1][M+1];
        for(int i=0;i<K;i++){
            str = scanner.nextLine();
            m = p.matcher(str);
            iii = -1;
            while(m.find()){
                iii++;
                if(iii>1)
                    break;
                t[iii] =new Integer(m.group()).intValue();
            }
            x = t[0];
            y = t[1];
            map[x][y] = 1;
        }
        int[][] book = new int[N+1][M+1];
        int max = 0;
        int s1,s2;
        for(int i=1;i<N-1;i++){
            for(int j=2;j<M;j++){
                s1 = 0;
                for(int k=-1;k<2;k++){//第一个镜子扫描
                    for(int l = -1 ; l < 2 ; l++){
                        s1 += map[i+k][j+l];
                        book[i+k][j+l] = -2;
                    }
                }
                for(int ii = 1 ; ii < N-1 ; ii++){
                    for(int jj = 2 ; jj < M ;jj++){
                        if(ii != i || jj != j){
                            s2 = 0;
                            for(int k = -1 ; k < 2 ; k++){
                                for(int l = -1 ;l < 2 ;l++){
                                    if(book[ii+k][jj+l] == 0){
                                        s2 += map[ii+k][jj+l];
                                    }
                                }
                            }
                            if((s1 + s2) > max){
                                max = s1 + s2;
                            }
                        }
                    }
                }
                for(int k=-1;k<2;k++){//第一个镜子扫描book复原
                    for(int l = -1 ; l < 2 ; l++){
                        book[i+k][j+l] = 0;
                    }
                }
            }
        }
        System.out.println(max);
        return;
    }
}

python

#-*- coding:utf-8 -*-
str = raw_input()
li = str.split(' ')
M = li[0]
N = li[1]
K = li[2]
M,N,K = int(M),int(N),int(K)

map = [i for i in range(M*N+1)]
book = [i for i in range(M*N+1)]
for i in range(0,N):
    for j in range(1,M+1):
        map[i*M+j] = 0
        
for i in range(K):
    str = raw_input()
    li = str.split(' ')
    x,y = li[0],li[1]
    x,y = int(x),int(y)
    map[(x-1)*M+y] = 1    


for i in range(1,M*N+1):
    book[i] = -1
    
max = 0
for i in range(1,N-1):
    for j in range(2,M):
        s1 = 0
        for k in range(-1,2):#第一个镜子扫描
            for l in range(-1,2):
                s1 += map[(i+k)*M+j+l]
                book[(i+k)*M+j+l] = -2 #表示访问过
        for ii in range(1,N-1):
            for jj in range(2,M):
                if ii*M+jj != i*M + j:
                    s2 = 0
                    for k in range(-1,2):#第二个镜子扫描
                        for l in range(-1,2):
                            if book[(ii+k)*M+jj+l] == -1:
                                s2 += map[(ii+k)*M+jj+l]
                    if s1 + s2 > max:
                        max = s1 + s2
                
        for t in range(1,M*N+1):
            book[t] = -1
            
print 'max:',max